25．（8分）如图，AB、AC分别与⊙O相切，切点分别为B、C，过点C作CD∥AB，交⊙
O于点D，连接BC、BD．
（1）判断BC与BD的数量关系，并说明理由； （2）若AB＝9，BC＝6，求⊙O的半径．
26．（10分）
实际情境
某中学九年级学生步行到郊外春游．一班的学生组成前队，速度为4 km/h，二班的学生组成后队，速度为6 km/h．前队出发1 h后，后队才出发，同时，后队派一名联络员骑自行车在两队之间不间断地来回进行联络，他骑车的速度为12 km/h． 数学研究
若不计队伍的长度，如图，折线A-B-C、A-D-E分别表示后队、联络员在行进过程中，离前队的路程．．．．．．y（km）与后队行进时间x（h）之间的部分函数图象． （1）求线段AB对应的函数关系式； （2）求点E的坐标，并说明它的实际意义；
（3）联络员从出发到他折返后第一次与后队相遇的过程中，当x为何值时，他离前队的
路程与他离后队的路程相等？
D O 1 21 3 24 C O D A B （第25题）
y/km A E B 2 C x/h （第26题）
27．（14分）如果两个多边形不仅相似（相似比不等于1），而且有一条公共边，那么就称这
两个多边形是共边相似多边形．例如，图①中，△ABC与△ACD是共AC边相似三角形，图②中，□ABCD与□CEFD是共CD边相似四边形．
B ①
C D A
F E A D A
②
C B C ③
B （第27题）
（1）判断下列命题的真假（在相应括号内填上“真”或“假”）：
①正三角形的共边相似三角形是正三角形．（ ▲ ）
②如果两个三角形是位似三角形，那么这两个三角形不可能是共边相似三角形．（ ▲ ）
（2）如图③，在△ABC中，∠C＝90°，∠A＝50°，画2个不全等的三角形，使这2个．．．．．
三角形均是与△ABC共BC边的相似三角形．（要求：画图工具不限，不写画法，保留画图痕迹或有必要的说明）
（3）图④是相邻两边长分别为a、b（a＞b）的矩形，图⑤是边长为c的菱形，图⑥是
两底长分别为d、e，腰长为f（0＜e－d＜2f）的等腰梯形，判断这三个图形是否存在共边相似四边形？如果存在，直接写出它们的共边相似四边形各边的长度．
b
a ④
c ⑤ （第27题）
f e ⑥ d
（4）根据（1）、（2）和（3）中获得的经验回答：如果一个多边形存在它的共边相似多
边形，那么它必须满足条件： ▲ ．
2013/2014学年度第二学期第一阶段学业质量监测试卷
九年级数学参考答案及评分标准
说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．
一、选择题（每小题2分，共计12分）
题号 答案 1 C 2 C 3 D 4 B 5 A 6 D 二、填空题（每小题2分，共计20分）
7．2 8．x≠－1 9．相交 10．2 11．a－b＝a＋(－b)（a、b是实数）
12．3 13．（4，6） 14．与它不相邻的三个内角的和减去180° 15．a4－4a3b＋6a2b2－4ab3＋b4 16．m＝n 三、解答题（本大题共11小题，共计88分） 17．（本题6分）
解法一：移项，得2x2－4x＝－1． ················································································· 1分
1
两边同时除以2，得x2－2x＝－．
2
1
配方，得 x2－2x＋1＝－＋1， ········································································· 2分
2
1
(x－1)2 ＝． ··················································································· 3分
2
由此可得 x－1＝±x1＝1＋
2． ·················································································· 4分 2
22，x2＝1－． ················································································· 6分 22
解法二：a＝2，b＝－4，c＝1．
b2－4ac＝(－4)2－4×2×1＝8＞0． ··································································· 2分 4±8
x＝ ···································································································· 4分
4＝1±
2． 2
22，x2＝1－． ················································································· 6分 22
x1＝1＋
18．（本题6分）
x2
解：(1－)÷2 x＋2x－4
(x＋2)－xx2－4＝· ······························································································· 2分
2x＋2＝
(x＋2) (x－2)2
· ···························································································· 4分
2x＋2
＝x－2． ············································································································ 6分
19．（本题6分）
x－42x＋1
解：根据题意，得＜． ··················································································· 1分
32
去分母，得2(x－4)＜3(2x＋1) ． ··········································································· 2分
去括号，得2x－8＜6x＋3． ·················································································· 3分 移项、合并同类项，得－4x＜11．·········································································· 5分
11系数化为1，得x＞－． ···················································································· 6分
4
20．（本题6分）
（1）证明：∵CD＝CE，AC＝CB，∠DCE＝∠ACB＝90°，
∴△ACE≌△BCD． ··················································································· 2分
∴AE＝BD． ····························································································· 4分
（2）解：45°或225°． ································································································ 6分 21．（本题8分）
解：（1）记“妈妈，我爱您”为a、“祝妈妈永远幸福”为b、“世上只有妈妈好”
为c以及“谢谢妈妈养育之恩”为d．任意刮开该版上的1枚邮票的丝带覆盖层，所有可能出现的结果有：a、a、b、b、c、c、d、d，共有8种，它们出现的可能性相同．所有的结果中，恰好均是 “妈妈，我爱您” 这句话（记
1
为事件A）的结果有2种，所以P（A）＝． ····················································· 4分
4（说明：没有说明等可能性扣1分．）
1
（2）． ······································································································ 8分
28
22．（本题8分）
解：（1）正确制作复合条形统计图或折线统计图． ···························································· 4分 （2）甲厂表述中的“平均”的含义是指该厂产品所测数据的平均数． ······························· 6分 乙厂表述中的“平均”的含义是指该厂产品所测数据的中位数． ······························· 8分
23．（本题8分）
解：如图，延长AD交FG于点E． ····································· 1分
FGFG
在Rt△FCG中，tanβ＝，∴CG＝． ······················ 2分
CGtanβFEFE
在Rt△FAE中，tanα＝，∴AE＝． ······················· 3分
AEtanα∵AE－CG＝AE－DE＝AD， ∴
FEFG
－＝AD． ··················································· 5分 tanαtanβ
α A D β
B C
G （第23题）
E F
FG－CDFG
即－＝AD．
tanαtanβ
AD·tanα·tanβ＋CD·tanβ∴FG＝＝115.5≈116． ························································· 7分
tanβ－tanα答：该信号发射塔顶端到地面的高度FG约是116 m． ···················································· 8分
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